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Force, Find the Angle Given 2 Forces of Equal Magnitude, but Different Direction

by Simon777
Tags: angle, direction, equal, force, forces, magnitude
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Simon777
#1
Sep14-11, 05:29 PM
P: 35
1. The problem statement, all variables and given/known data
Two horses pull horizontally on ropes attached to a tree stump. Each horse pulls with a force of magnitude F. If the resultant force (vector R) has the magnitude R = 1.51 F, what is the angle (in degrees) between the two ropes?


2. Relevant equations



3. The attempt at a solution
Ry= 0+cos
Rx= 1+ sin

R= 1.51

sin=cos/1.54
tan= 1/1.54
angle= 33 degrees

33 * 2 = total angle for both of 66 degrees
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Simon777
#2
Sep14-11, 05:42 PM
P: 35
I'm on my 5th hour for this problem and still can't get it. I realize it's necessary to assign an arbitrary value of say 1 to F. That gives F=1 and R=1.51, but everything I've tried to do with that information has failed.
Allenman
#3
Sep14-11, 06:02 PM
P: 58
Using vector addition... We don't know whether the three vectors make a right triangle (when putting the two F's nose to tail), so it's easier to divide them into 2 triangles. We'll call the new segment G.

So now F becomes the Hypotenuse, G is Opposite, and 1.51/2 F is the adjacent (divided in half because it's now the adjacent on 2 triangles instead of just 1).

F sin(angle) = 1.51/2 F => divide out the F => arcsin(1.51/2) = angle = 49.025 degrees

That's just half of your angle though since you split it into two triangles
49.025*2 = 98.05 degrees

Simon777
#4
Sep14-11, 06:15 PM
P: 35
Force, Find the Angle Given 2 Forces of Equal Magnitude, but Different Direction

Quote Quote by Allenman View Post
Using vector addition... We don't know whether the three vectors make a right triangle (when putting the two F's nose to tail), so it's easier to divide them into 2 triangles. We'll call the new segment G.

So now F becomes the Hypotenuse, G is Opposite, and 1.51/2 F is the adjacent (divided in half because it's now the adjacent on 2 triangles instead of just 1).

F sin(angle) = 1.51/2 F => divide out the F => arcsin(1.51/2) = angle = 49.025 degrees

That's just half of your angle though since you split it into two triangles
49.025*2 = 98.05 degrees
The professor gave an answer of 81.9 degrees.
Allenman
#5
Sep14-11, 06:25 PM
P: 58
Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.
Simon777
#6
Sep14-11, 06:32 PM
P: 35
Quote Quote by Allenman View Post
Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.
I really appreciate the response, I initially did what you did and was told to rethink it and that it was 81.9. That's what's been making me go insane. I don't know why this method is wrong.
Simon777
#7
Sep14-11, 07:35 PM
P: 35
Quote Quote by Allenman View Post
Hmmmm maybe someone else will chime in... I don't see anything wrong with what I did, maybe I'm overlooking something.
You had it right and explained it right, but used arcsin instead of arccos. I get how to do this now, thank you so much for helping me finally understand this.


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